Here's the question you clicked on:
ilikephysics2
Find the first 4 terms of the taylor series of sin x at x = pi/4.
The first 4 derivatives are cos(x) then -sin(x) then -cos(x) and then sin(x)
sin(pi/4) + cos(pi/4) * (x-pi/4) -sin(pi/4) * (1/2!) (x-pi/4)^2 .....
pretty straightforward.. what was your question about?
that's the first 3 terms... you should be able to get the 4th now?
That does not look right though
well... I lied a bit... a few of those terms are zero...
but you can work from them... right?
f(x) = f(x0) + (x-x0)*f'(x0)/1! + (x-x0)^2*f"(x0)/2! + (x-x0)^3*f'"(x0)/3!
with x0 = pi/4, for f(x) = sinx --> f(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'(x) = cosx --> f'(pi/4) = cos(pi/4) = 1/2*sqrt(2) f"(x) = -sinx --> f"(pi/4) = -sin(pi/4) = -1/2*sqrt(2) f'"(x) = -cosx --> f'"(pi/4) = -cos(pi/4) = -1/2*sqrt(2) now, just subtitute all numbers above to generally formula !!!